Help with 70-291 Question
HowardM
Member Posts: 5 ■□□□□□□□□□
I ran accross this question from the MCSA/MCSE 70-291 Self-Paced Training Kit. I am not sure why the correct answer is correct. Can anyone give me an in depth explanation to how they arrived at this answer on this question?
You work for a large organization that has leased the Class B
address 131.188.0.0. It uses VLSM. The network is divided into 6
backbone subnets, each with 8190 host addresses. These hosts are
then subnetted as required. The 131.188.96.0/19 subnet is divided
into 62 equal-sized subnets. On which subnet would you find host
131.188.97.140?
1. 131.188.97.128/25 <Correct>
2. 131.188.96.0/19
3. 131.188.97.0/25
4. 131.188.97.0/24
Explanation :
The host address range for this subnet is 131.188.97.129 through
131.188.97.254. It is also the most specific subnet in the path.
Host 131.188.97.140 is found on this subnet.
Objective:
Implementing, Managing, and Maintaining IP Addressing
Sub Objective(s):
Configure TCP/IP addressing on a server computer
:
You work for a large organization that has leased the Class B
address 131.188.0.0. It uses VLSM. The network is divided into 6
backbone subnets, each with 8190 host addresses. These hosts are
then subnetted as required. The 131.188.96.0/19 subnet is divided
into 62 equal-sized subnets. On which subnet would you find host
131.188.97.140?
1. 131.188.97.128/25 <Correct>
2. 131.188.96.0/19
3. 131.188.97.0/25
4. 131.188.97.0/24
Explanation :
The host address range for this subnet is 131.188.97.129 through
131.188.97.254. It is also the most specific subnet in the path.
Host 131.188.97.140 is found on this subnet.
Objective:
Implementing, Managing, and Maintaining IP Addressing
Sub Objective(s):
Configure TCP/IP addressing on a server computer
:
Comments
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eastp Member Posts: 179the "/25" indicates the first 25 bits are used to identify the unique network
(11111111.11111111.11111111.10000000)
leaving the remaining bits to identify the specific host.
(The remaining bits can add up to 1111111 = 127)
so if you take the .128 + 127 = 255 -1(255 is used as broadcast) = 254
Meaning that all valid hosts need to be in the range of 131.188.97.129 to 131.188.97.254
Kind regards.
EastpMultitasking:
Screwing up several things at once.